1) An ? particle enters a constant magnetic field as shown in the diagram. The m

Question

1) An ? particle enters a constant magnetic field as shown in the diagram. The magnetic field is directed into the screen and has a magnitudeB = 1.52 T. The initial velocity of the ? particle is v = 1.84 × 105 m/s directed up the screen.

What is the size of the force experience by the ? particle?

2)

What is the direction of this force?

Select one:

To the right

To the left

Up the screen

Out of the screen

It is zero so it has no direction

Down the screen

Into the screen

3)

What is the magnitude of the acceleration experienced by the ? particle?

4)

This force causes the ? particle to undergo circular motion. What is the radius of the path it follows?

Explanation / Answer

1)   force experience by the particle = qvB

                                                              = 2*1.6*10-19*1.84 × 105*1.52

                                                                 = 8.949*10-14 N

2) Direction of force - to the left

3) magnitude of the acceleration experienced = 8.949*10-14/6.64424* 10-27

                                                                                                           = 1.3468*1013 m/sec2

4) Here, mv2/r = 8.949*10-14

               => r = 6.64424* 10-27*1.84 × 105*1.84 × 105/8.949*10-14

                              = 2.513*10-3 m

                      = 0.2513 cm   ------------------------> radius of alpha particle

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